Answer
(a) $v_B = \frac{m_A~v_A}{m_B}$
(b) $\frac{K_A}{K_B} = \frac{m_B}{m_A}$
Work Step by Step
(a) Since the initial momentum is zero, the momentum of the two pieces will be equal in magnitude.
$m_B~v_B = m_A~v_A$
$v_B = \frac{m_A~v_A}{m_B}$
(b) $\frac{K_A}{K_B} = \frac{\frac{1}{2}m_A~v_A^2}{\frac{1}{2}m_B~v_B^2}$
$\frac{K_A}{K_B} = \frac{m_A~v_A^2}{m_B~(\frac{m_A~v_A}{m_B})^2}$
$\frac{K_A}{K_B} = \frac{m_A~v_A^2~m_B^2}{m_B~m_A^2~v_A^2}$
$\frac{K_A}{K_B} = \frac{m_B}{m_A}$
